Reference: LeetCode
Difficulty: Easy

Problem

Given two arrays, write a function to compute their intersection.

Note:

  • Each element in the result must be unique.
  • The result can be in any order.

Example: 

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Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]

Follow-Up: If it is sorted, can we do it without extra space?

Analysis

Brute-Force

Use one set for the result list to avoid duplicates.

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public int[] intersection(int[] nums1, int[] nums2) {
  Set<Integer> set = new HashSet<>();

  for (int val1 : nums1) {
    for (int val2 : nums2) {
      if (val1 == val2) {
        set.add(val1);
      }
    }
  }
  // to array
  int[] result = new int[set.size()];
  int count = 0;
  for (int val : set) {
    result[count] = val;
    count += 1;
  }
  return result; 
}

Time: $O(M \times N)$
Space: $O(M + N)$

Use Extra Sets

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public int[] intersection(int[] nums1, int[] nums2) {
  Set<Integer> interSet = new HashSet<>(); // use hash set to avoid duplicates
  Set<Integer> set = new HashSet<>();
  // add nums1 to set
  for (int val : nums1) set.add(val);
  // check if nums2 in set
  for (int val : nums2) {
    if (set.contains(val)) {
      interSet.add(val);
    }
  }
  // to array
  int[] result = new int[interSet.size()];
  int count = 0;
  for (int val : interSet) {
    result[count] = val;
    ++count;
  }
  return result;
}

Solution:

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public int[] intersection(int[] nums1, int[] nums2) {
  Set<Integer> set1 = new HashSet<>();
  Set<Integer> set2 = new HashSet<>();
  for (int val : nums1) set1.add(val);
  for (int val : nums2) set2.add(val);
  if (set1.size() < set2.size()) {
    return intersectionHelper(set1, set2);
  } else {
    return intersectionHelper(set2, set1);
  }
}

private int[] intersectionHelper(Set<Integer> set1, Set<Integer> set2) {
  // assume set1.size() < set2.size()
  int[] result = new int[set1.size()];
  int count = 0;
  for (int val : set1) {
    if (set2.contains(val)) {
      result[count] = val;
      count += 1;
    }
  }
  return Arrays.copyOf(result, count); // init with count-element array
}

Time: $O(M + N)$
Space: $O(M + N)$

Built-In Intersection

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public int[] intersection(int[] nums1, int[] nums2) {
  Set<Integer> set1 = new HashSet<>();
  Set<Integer> set2 = new HashSet<>();
  for (int val : nums1) set1.add(val);
  for (int val : nums2) set2.add(val);
  set1.retainAll(set2);
  int[] result = new int[set1.size()];
  int count = 0;
  for (int val : set1) {
    result[count++] = val;
  }
  return result;
}

Time: $O(N + M)$
Space: $O(N + M)$

Two Pointers If Sorted

Note: Be careful of the conditions.

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public int[] intersection(int[] nums1, int[] nums2) {
  // Assume they are sorted
  Arrays.sort(nums1);
  Arrays.sort(nums2);
  
  Set<Integer> result = new HashSet<>();
  
  int n1 = nums1.length, n2 = nums2.length;
  int i1 = 0, i2 = 0;
  while (i1 < n1 && i2 < n2) {
    while (i1 < n1 && i2 < n2 && nums1[i1] < nums2[i2]) ++i1;
    while (i1 < n1 && i2 < n2 && nums1[i1] > nums2[i2]) ++i2;
    
    // critical: consider nums1[i1] < nums2[i2] --> if it is, we should continue to the next loop
    if (i1 < n1 && i2 < n2 && nums1[i1] == nums2[i2]) {
      result.add(nums1[i1]); // we use hash set --> no need to handle duplicates
      ++i1;
      ++i2; // update indices should be inside the if logic
    }
  }
  
  int[] output = new int[result.size()];
  int i = 0;
  for (int val : result) {
    output[i] = val;
    ++i;
  }
  
  return output;
}